From the given fig tanα = 3/4,

cosα = 4/5 & sinα = 3/5,

Consider equilibrium of block W

RV = 0; R_{2} = Wcosα ...(1)

RH = 0; T_{1} = µR_{2} + Wsinα ...(2)

Putting the value of equation(1) in (2)

T_{1} = µWcos α + Wsinα

= 0.3 X W(4/5) + W(3/5)

T_{1} = 0.84W ...(3)

For pulley; T_{2}/T1 = e^{µ}_{1}^{θ}

T_{2} = T1 X e^{µ}_{1}^{θ}

= 0.84We^{(0.2 X )}

T_{2} = 1.574W.....(4)

Consider equilibrium of block 100N

RV = 0; R_{1} = 100cos α + R_{2} ...(5)

R_{1} = 100cosα + Wcosα = 100(4/5) + W(4/5)

R_{1 }= 80 + 0.8W ...(6)

R_{H} = 0;

T_{2 }= 100sinα –µR_{1} -µR_{2}

T_{2} = 100(3/5) –0.3[(80 + 0.8W) – W(4/5)]

1.574W = 60 – 24 – 0.24W – 0.24W

**W = 17.53N.**